Tutorial06-solutions

# Of sin 2 sin 2 indep of sin 2 thus

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Unformatted text preview: a) First, if B − A ∈ Z then for any α we have S B B cos 2π ( + α) 1 [cos(2πA+2πα)−cos(2πB+2πα)] =− = 2π 2π =A 1 [cos(2πA + 2πα) − cos(2πA + 2πα + 2π (B − A))] = 0 = 2π sin 2π ( + α) A ∈Z since cos has period 2π . On the other hand, suppose B − A ∈ Z. Then there exists an α, that is α = −A for which we get, using the same calculation, B sin 2π ( + α) A == 1 [cos(2πA + 2πα) − cos(2πA + 2πα + 2π (B − A))] = 2π 1 1 [cos 0 − cos 2π (B − A)] = [1 − cos 2π (B − A)] = 0 2π 2π ∈Z =1 (b) We have [ ×[ sin 2π ( + α) · sin 2π ( + β) = sin 2π ( + α) · sin 2π ( + β) = indep. of sin 2π ( + β) sin 2π ( + α) · = indep. of sin 2π ( + β) · Thus, using the result from part (a), we see that we have and only if [ ×[ sin 2π ( + α) · sin 2π ( + β) for any α β ∈ R. 10 sin 2π ( + α) − ∈ Z or =0 − ∈ Z if − (c) Since by assumption, [ ∈ Z or − ∈ Z, part (b) tells us that sin 2π ( + α) · sin 2π ( + β) ×[ for all α, β. (d) Since the small rectangles [×[ don’t overlap and their union is the big rectangle [×[, we have by the additivity of integration and by part (c) that [ ×[ sin 2π ( + α) · sin 2π ( + β) = =1 [ ×[ sin 2π ( + α) · sin 2π ( + β) =0 for all α β ∈ R. (e) Using part (d) together with part (b) again, tells us that E − ∈ Z or − ∈ Z. :) Evaluate...
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## This document was uploaded on 02/10/2014.

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